If $\mathbb V$ is a vector space over $\mathbb F$ then does $\mathbb V$ contain $\mathbb F$ I know that if $\Bbb{K}$ is an extension field of $\Bbb{F}$ then $\Bbb{K}$ is also a vector space over $\Bbb{F}$. However, I want to know, given any vector space V over a field (not necessarily finite) $\Bbb{F}$, is it necessary that V should always contain $\Bbb{F}$. [I want a convincing explanation of this and also an example if this is not True]. for example, I know of the most trivial case where $\Bbb{F} = \Bbb{R}$ and $V = \Bbb{R}^n$ in which case $V$ contains $\Bbb{F}$
Thanks in advance
 A: Define "contain". You say that $\mathbb{R}^n$ contains $\mathbb{R}$ but that is not the case, at least not literally contain. The $\mathbb{R}^n$ for $n>1$ contains $n$-tuples of reals, which are not real numbers, and so no inclusion can ever happen here.
$\mathbb{R}$ does however embed as a linear subspace into $\mathbb{R}^n$ via for example $r\mapsto (r,0,\ldots,0)$. This is especially important when considering things like complex numbers $\mathbb{C}$ which by definition is $\mathbb{R}^2$ with additional multiplication. The standard way of treating $\mathbb{R}$ as a subset of $\mathbb{C}$ is precisely via $r\mapsto (r,0)$ embedding. This embedding has a nice additional property that it preserves multiplication, and so $\mathbb{R}$ embeds into $\mathbb{C}$ as a subfield. Note however that this is not the only way of embedding $\mathbb{R}$ into $\mathbb{C}$, even as a subfield, see this.
More generally let $V$ be any nonzero vector space over any field $k$ and pick any nonzero vector $v\in V$. Define $kv=\{\lambda v\ |\ \lambda\in k\}$. Clearly $kv$ is a vector subspace of $V$ and moreover there's a linear embedding of $k$ into $V$ via $\lambda\mapsto \lambda v$ whose image is $kv$.
Therefore $k$ can be linearly embedded into any nonzero $k$-vector space, in multiple ways. Note that we cannot expect any more than "linear" embedding, since we don't have (vector) multiplication on $V$.
